what is the enthalpy of combustion when 1 mol c6h6(g) completely reacts with oxygen? 2c6h6(g) + 15o2(g) →…

what is the enthalpy of combustion when 1 mol c6h6(g) completely reacts with oxygen? 2c6h6(g) + 15o2(g) → 12co2(g) + 6h2o(g) -6339 kj/mol -3169 kj/mol 1268 kj/mol 6339 kj/mol done compound δhf (kj/mol) c6h6(g) 82.90 co2(g) -393.50 h2o(g) -241.82

what is the enthalpy of combustion when 1 mol c6h6(g) completely reacts with oxygen? 2c6h6(g) + 15o2(g) → 12co2(g) + 6h2o(g) -6339 kj/mol -3169 kj/mol 1268 kj/mol 6339 kj/mol done compound δhf (kj/mol) c6h6(g) 82.90 co2(g) -393.50 h2o(g) -241.82

Answer

Explanation:

Step1: Recall the formula for enthalpy of reaction

$\Delta H_{rxn}=\sum n_p\Delta H_f(products)-\sum n_r\Delta H_f(reactants)$

Step2: Identify reactants and products and their coefficients

For the reaction $2C_6H_6(g)+15O_2(g)\rightarrow12CO_2(g) + 6H_2O(g)$, $n_{C_6H_6}=2$, $n_{O_2} = 15$, $n_{CO_2}=12$, $n_{H_2O}=6$. The $\Delta H_f$ of $O_2(g)$ is 0 kJ/mol.

Step3: Calculate enthalpy of reaction for 2 moles of $C_6H_6$

$\Delta H_{rxn}=[12\times(- 393.50)+6\times(-241.82)]-[2\times82.90 + 15\times0]$ $=[-4722-1450.92]-[165.8]$ $=-4722-1450.92 - 165.8=-6338.72\approx - 6339$ kJ/mol for 2 moles of $C_6H_6$.

Step4: Calculate enthalpy of combustion for 1 mole of $C_6H_6$

Since the calculated $\Delta H_{rxn}$ is for 2 moles of $C_6H_6$, for 1 mole of $C_6H_6$, $\Delta H=\frac{-6339}{2}=-3169.5\approx - 3169$ kJ/mol

Answer:

-3169 kJ/mol